Problem 86 - Essential Calculus

Question

The polar form of an equation for a curve is \(r=f(\sin \theta)\). Show that the form becomes (a) \(r=f(-\cos \theta)\) if the curve is rotated counterclockwise \(\pi / 2\) radians about the pole. (b) \(r=f(-\sin \theta)\) if the curve is rotated counterclockwise \(\pi\) radians about the pole. (c) \(r=f(\cos \theta)\) if the curve is rotated counterclockwise \(3 \pi / 2\) radians about the pole.

Step-by-Step Solution

Answer

When we rotate the curve counterclockwise by \(\pi / 2\) radians, the polar equation changes to \(r=f(-\cos \theta)\). When we rotate by \(\pi\) radians, it changes to \(r=f(-\sin \theta)\). And when we rotate by \(3 \pi / 2\) radians, it becomes \(r=f(\cos \theta)\).

Step 1: Rotation By \(\pi / 2\) radians counterclockwise

When we rotate a curve counterclockwise by \(\pi / 2\) radians, we replace \(\theta\) in our given polar equation with \(\theta - \pi / 2\). Because \(\sin(\theta - \pi / 2) = -\cos(\theta)\), the equation \(r = f(\sin \theta)\) becomes \(r = f(-\cos \theta)\), which is what we needed to show.

Step 2: Rotation By \(\pi\) radians counterclockwise

When we rotate a curve counterclockwise by \(\pi\) radians, we replace \(\theta\) in our given polar equation with \(\theta - \pi\). Because \(\sin(\theta - \pi) = -\sin(\theta)\), the equation \(r = f(\sin \theta)\) becomes \(r = f(-\sin \theta)\), which is what we needed to show.

Step 3: Rotation By \(3 \pi / 2\) radians counterclockwise

When we rotate a curve counterclockwise by \(3 \pi / 2\) radians, we replace \(\theta\) in our given polar equation with \(\theta - 3 \pi / 2\). Because \(\sin(\theta - 3 \pi / 2) = \cos(\theta)\), the equation \(r = f(\sin \theta)\) becomes \(r = f(\cos \theta)\), which is what we needed to show.